Aircraft propulsion 1e saeed farokhi solutions manual

Aircraft propulsion 1e saeed farokhi solutions manual

http://www.mediafire.com/view/lfazyasgagssoq3/ch02.doc

Chapter 2 Solutions

Problem 2.1

Given that:

Solution:

a) Using Equation 2.102-a,

Rearranging and solving for u_1,

In order to get M₁:

 

b) First, we need to find M₂ of the flow

Refer to Appendix C, Normal Shock Table (g=1.4),

 for M₁=1.972

Then, with the help of Equation 2.104,

Problem 2.2

 

Solution

a) From the figure shown, an oblique shock forms between region 1 and 2:

For M₁ = 3.0, q = 5^o

According to Figure 2.23(a), b = 23^o

Using the normal shock table,

For M_1n = 1.172,

 and    

b) From the figure shown, and oblique shock forms between region 1 and 3:

For M₁ = 3.0, q = 10^o

According to Figure 2.23(a), b = 27.5^o

Using the normal shock table,

For M_1n = 1.385,

 and    

c)

From the figure,

The length of the face normal to the p₂ and p₃ is:

 

We can write,

Note:

Therefore,

Problem 2.3

Solution:

An oblique shock occurs at the concave corner between station 1 and 2:

M₁ = 2

q = 10^o

From Figure 2.23(a), b = 39^o

Using the normal shock table,

For M_1n = 1.26,  and

We then solve for p₂,

Also,

From the isentropic table, using M₂ = 1.665:

The Pitot tube measures the stagnation pressure in region 3 to be:

Problem 2.4

Solution:

An oblique shock occurs at the concave corner between station 1 and 2:

b₁ = 30^o

q₁ = 9^o

From Figure 2.23(a), M₁ = 2.6

Using the normal shock table,

For M_1n = 1.3,  

A reflected oblique shock occurs at the concave corner between station 2 and 3:

M₂ = 2.193

q = 9^o

From Figure 2.23(a), b = 36.5^o

Using the normal shock table,

For M_2n = 1.305,  

Problem 2.5

Solution:

a) An oblique shock occurs between the test section and region 2.

 Flow has to turn 10^o across the oblique shock and p₂ can be determined by:

M_T.S. = 2.0, q₁ = 10^o; According to Figure 2.23(a), b₁ = 39^o

Using the normal shock table,

For M_1n = 1.259,   and

Referring to isentropic flow table,

For

Therefore,  

b) From the Prandtl-Meyer (P-M) Table,

For M_T.S.=2.0, n₁(M₁) = 26.38^o

n₃(M₃) = n₁(M₁) + q = 31.38^o

Using n₃(M₃) = 31.38^o, we return to the P-M Table to get

Referring to isentropic flow table,

For

Flow across expansion wave is isentropic,

Therefore,

c) The flow between region 2 and 4 has to travel across a series of expansion waves:

Referring to isentropic flow table,

For

From the Prandtl-Meyer (P-M) Table,

For M₂=1.665, n₂(M₂) = 16.928^o

n₄(M₄) = n₂(M₂) + q₂ = 16.928^o + 10^o = 26.928^o

Using n₄(M₄) = 26.928^o, we return to the P-M Table to get

Referring to isentropic flow table,

For

Flow across expansion wave is isentropic,

Therefore,

Problem 2.6:

Solution:

First, we need to establish the wall turning angle, n₁(M₁)

From the Prandtl-Meyer (P-M) Table,

For M₁=2.0, n₁(M₁) = 26.38^o

a) n₂(M₂) = n₁(M₁) + q = 26.38^o + 30^o = 56.38^o

Using n₂(M₂) = 56.38^o, we return to the P-M Table to get

b) Using isentropic relations, we determine  for the flow in regions 1 and 2,

Referring to isentropic flow table,

M₁ = 2.0   and

Therefore,

c) The fan angle, d can be determined by identifying the wave angle for 1 and 2:

From P-M table, M₁ = 2.0   and

Therefore,  

Problem 2.7

Solution:

First, we state regions 2 and 3 to be the top and region 4 of the half-diamond airfoil.

Region 1 is the free stream condition.

We must note that a normal shock exists ahead of the Pitot tube due to supersonic flow over a blunt body. Pitot tube measures total or stagnation pressure of the flow.

No shock exists between region 1 and 4. Thus, p₄=p₁.

From the Normal Shock table,

For M₁=3.0,

Pitot tube in region 4 measures:

An oblique shock exists between region 1 and 2.

M₁ = 3.0

q₁ = 30^o

From Figure 2.23(a), b₁ = 52^o

Using the normal shock table,

For M_1n = 2.364, and

For M₂=1.408,   from normal shock table.

Pitot tube in region 2 measures:

The flow then flows across an expansion wave from region 2 to 3. Thus, we need to establish the wall turning angle, n₂(M₂)

From the Prandtl-Meyer (P-M) Table,

For M₂=1.408, n₂(M₂) = 8.99^o; q₂=60^o

n₃(M₃) = n₂(M₂) + q₂ = 68.99^o

Using n₃(M₃) = 68.99^o, we return to the P-M Table to get

Flow across expansion wave is isentropic,

Using the normal shock table,

For M₃ = 4.3, and

Pitot tube in region 3 measures:

Problem 2.8

Solution:

a) An expansion fan occurs between region 1 and 2 while an oblique shock turn flow into itself between region 1 and 3.

Region 1 and 2:

q=30^o; From the Prandtl-Meyer (P-M) Table,

For M₁=5.0, n₁(M₁) = 76.92^o. Thus, n₂(M₂) = n₁(M₁) + q = 106.92^o

Using n₂(M₂) = 106.92^o, we return to the P-M Table to get

Referring to isentropic flow table,

For  and

Flow across expansion wave is isentropic, p_t,2=p_t,1,

Region 1 and 3:

For M₁ = 5.0, q = 30^o; According to Figure 2.23(a), b = 42^o

Using the normal shock table,

For M_1n = 3.345,   and  

  and

, also

b)

Problem 2.9

Solution:

a)

Eqn 2.104 states:

Using the relation,

b) Prandtl’s Relation:

c) Using Equation 2.102-a,  

d) According to conservation of energy across normal shock,

Problem 2.10

Assuming negligible heat and pressure losses through the wind tunnel,

p_room=p_t,room=p_t,T.S.=100 kPa and T_room=T_t,room=T_t,T.S..= 288.13 K

Solution:

a)

b)

Force on each glass window:

Problem 2.11

Solution:

Given: q=30^o; p_base=p_∞; M_∞=3.0 and region 1 is downstream of the oblique shock.

(b/2)=a sinq , where a is the length of face normal to p₁ and p₂. 

 

According to Figure 2.23(a), b = 52^o

Using the normal shock table,

For M_∞,n = 2.364,   

Therefore,

 

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Aircraft propulsion 1e saeed farokhi solutions manual